Kristen invests money in a bank and makes no additional deposits or withdrawals. the bank pays 1.4% interest compounded annually. to the nearest tenth of a year, how long must she leave the money in the bank for it to double? solve algebraically.

approximately 50 years.

Step-by-step explanation:

Let $P represent the initial amount that she deposited. It means that principal,

P = $P

It was compounded annually. This means that it was compounded once in a year. So

n = 1

The rate at which the principal was compounded is 1.4%. So

r = 1.4/100 = 0.014

The formula for compound interest is

A = P(1+r/n)^nt

A = total amount in the account at the end of t years. For the initial amount to double, it means that

A = 2P

Therefore

2P = P (1+0.014/1)^1×t

2P/P = (1.014)^t

2 = (1.014)^t

Taking log to base 10 of both sides, it becomes

Log 2 = log 1.014^t

Log 2 = tlog 1.014

0.301 = 0.006t

t = 0.301/0.006 = 50.2 years

51.4 years

Step-by-step explanation:

There is what we call rule of 72 in Accounting to find the time  a sum of money will double if it is compounded annually at any given rate . The formula is given as :

t =

where t is the time and r is the rate.

t = ?

r = 1.4

Therefore :

t = 72/1.4

t = 51.4286

to the nearest tenth

t = 51. 4 years

"image"

step-by-step explanation:

in most of these problems, the notation is that the point with a "prime" is the image of the point without. that is, b' is the image of b.

here, there is no dilation or reflection. there is simply translation (with no direction shown). which figure you call the "preimage" and which one you call the "image" is up to you. as a matter of convention, the naming is as described above.